What is the domain and range of #f(x)=(x^2+1)/(x+1)#?

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Answer 1 · 2017-09-04T17:42:53

The domain is #x in (-oo,-1) uu(-1,+oo)#
The range is #y in (-oo, -2-sqrt8] uu [-2+sqrt8, +oo)#

As we cannot divide by #0#, #x!=-1#

The domain is #x in (-oo,-1) uu(-1,+oo)#

Let #y=(x^2+1)/(x+1)#

So,

#y(x+1)=x^2+1#

#x^2+yx+1-y=0#

In order for this equation to have solutions, the discriminant is

#Delta<=0#

#Delta=y^2-4(1-y)=y^2+4y-4>=0#

#y=(-4+-(16-4*(-4)))/(2)#

#y=(-4+-sqrt32)/2=(-2+-sqrt8)#

#y_1=-2-sqrt8#

#y_2=-2+sqrt8#

Therefore the range is

#y in (-oo, -2-sqrt8] uu [-2+sqrt8, +oo)#

graph{(x^2+1)/(x+1) [-25.65, 25.66, -12.83, 12.84]}